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Question

A particle is projected at $${ 60 }^{ }$$ to the horizontal with a kinetic energy KE. The kinetic energy at the highest point is :


A
KE
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B
Zero
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C
KE4
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D
KE2
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Solution

The correct option is C $$\dfrac { KE }{ 4 } $$
Let initial velocity of the particle be $$v$$.
$$\therefore KE=\dfrac { 1 }{ 2 } { mv }^{ 2 }$$
at the highest point of its flight
    $$v=u\cos { { 60 }^{ } } =\dfrac { u }{ 2 } $$
$$\therefore $$ Kinetic energy at this point
   $$=\dfrac { 1 }{ 2 } { mv }^{ 2 }=\dfrac { 1 }{ 2 } \dfrac { { mu }^{ 2 } }{ 4 } =\dfrac { KE }{ 4 } $$

Physics

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